imported>Patrick at 14:30, 12 August 2011

2011-08-12T14:30:01Z

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We start by drawing a free-body diagram for the forces acting on the box: 
[[image:Helena_Work-Energy_Ex_2_Soln.png|TOP]] 

'''(a)''' The person exerts 80 N through 3 m in the direction of the displacement along the slope. The x-component of the force is <math>F_x = 80 \cos(0^{\circ}) = 80 N</math> and the x-component of the displacement is <math>\Delta x = 3 m</math>. The work done by the person is

<math>W_F = (80)(3) = 240 J</math>
 
'''(b)''' The gravitational force (100 N) makes an angle of -120° and so the x-component of the gravitational force is <math>F_{G_x} = 100 \cos(-120^{\circ}) = -50 N</math> and therefore the work done by gravity is

<math>W_{F_G} = -150 J</math>
 
'''(c)''' The force of friction is given to be 22 N. The x-component of the force of friction is <math>F_{f_x} = 22 \cos(-180^{\circ}) = -22 N</math>. Therefore the work done by friction is

<math>W_{F_f} = -66 J</math>
 
'''(d)''' The work done by the normal force is zero because the x-component of the normal force is zero. 

'''(e)''' The change of the kinetic energy is equal to the total work. To find the total work we have to add the work done by each individual force:

<math>\Delta K = W_{total} = W_F + W_{F_G} + W_{F_f} = 240 - 150 - 66 = 24 J</math>

Work-Energy Theorem EX 2 - Revision history

imported>Patrick at 14:30, 12 August 2011